def median_1(a, b):
# 思路一: 先組合成乙個有序數列,再取中位數
# 時間複雜度o(m+n)
len_a = len(a)
len_b = len(b)
c =
if len_a == len_b == 0:
raise valueerror
i = j = 0
for k in range(0, len_a + len_b):
if j == len_b or (i < len_a and a[i] <= b[j]):
i += 1
else:
j += 1
half = (len_a + len_b) // 2
if (len_a + len_b) % 2 == 0:
return (c[half - 1] + c[half]) / 2
else:
return c[half]
def median_2(a, b):
# 思路二: 沒有必要完全產生出第三個列表,我們在一開始就可以知道需要取的索引,且可以用變數記錄而不新建列表
# 時間複雜度: o((m+n)/2) => o(m+n)
len_a = len(a)
len_b = len(b)
if len_a == len_b == 0:
raise valueerror
half = (len_a + len_b) // 2 + 1
pre = cur = i = j = 0
for k in range(0, half):
if j == len_b or (i < len_a and a[i] <= b[j]):
pre = cur
cur = a[i]
i += 1
else:
pre = cur
cur = b[j]
j += 1
if (len_a + len_b) % 2 == 0:
return (pre + cur) / 2
else:
return cur
def median_3(a, b, k=none):
# 思路三: 求中位數的問題可以看作是求第(m+n)/2小的數的問題.如果是偶數個,則是第(m+n)/2小和第(m+n)/2+1小的均值.
# 時間複雜度: o(log(m+n))
m, n = len(a), len(b)
if m > n:
n, m, a, b = m, n, b, a
if n == 0:
raise valueerror
if k == none:
k1 = (m + n + 1) // 2
k2 = (m + n + 2) // 2
return (median_3(a, b, k1) + median_3(a, b, k2)) / 2
if m == 0:
return b[k - 1]
if k == 1:
return a[0] if a[0] <= b[0] else b[0]
half = k // 2
index_a = min(m - 1, half - 1)
index_b = min(n - 1, half - 1)
if a[index_a] <= b[index_b]:
return median_3(a[index_a + 1:], b, k - index_a - 1)
else:
return median_3(a, b[index_b + 1:], k - index_b - 1)
def median_4(a, b):
# 思路四:二分法, i = 0 ~ m, j = (m + n + 1) / 2 - i, 需保證j>=0, 即n>=m
# 時間複雜度: o(log(min(m,n)))
m, n = len(a), len(b)
if m > n:
m, n, a, b = n, m, b, a
if n == 0:
raise valueerror
i_min = 0
i_max = m
half = (m + n + 1) // 2
while i_min <= i_max:
i = (i_min + i_max) // 2
j = half - i
if i > 0 and a[i - 1] > b[j]:
# i太大了
i_max = i - 1
elif i < m and a[i] < b[j - 1]:
# i太小了
i_min = i + 1
else:
if i == 0:
max_of_left = b[half - 1]
elif j == 0:
max_of_left = a[half - 1]
else:
max_of_left = max(a[i - 1], b[j - 1])
if (m + n) % 2 == 1:
return max_of_left
if i == m:
min_of_right = b[j]
elif j == n:
min_of_right = a[i]
else:
min_of_right = min(a[i], b[j])
return (max_of_left + min_of_right) / 2
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